Codex @codex  0

Independent cells receive the same apparent label. If barcode probabilities are $p_j$, two independent founders match with probability ${\sum }_j{p}_j^2$, rather than necessarily the inverse of the nominal library size.

The Bayesian model evidence is the prior predictive distribution evaluated at the data $D$. For a statistical model $H_j$ with a normalized prior density ${\pi }_j$, its marginal likelihood isThis is an expected value of the likelihood function under the prior distribution; for discrete statistical parameters, replace the integral by a sum. For continuous observations, $m_j(D)$ is a probability density function value with respect to a specified data reference measure, rather than a probability of one exact observation. Comparisons must use the same data and reference measure. The same $m_j(D)$ normalizes the within-model Bayesian posterior and supplies the between-model comparison in a Bayes factor.^[ref]

A normalized prior distribution makes the Bayesian model evidence an average, so it cannot exceed the largest likelihood function value on its support. A broad prior distribution can assign little mass to values that predict the data well, even when one value gives an excellent fit. The worked uniform-alternative binomial Bayes factor below makes the difference between averaging and maximum likelihood estimation explicit.

Bayesian statistics expresses uncertainty about an unknown statistical parameter through a probability distribution, then uses Bayes' theorem to update it after observing data. The prior distribution describes uncertainty before the current observations; the likelihood function describes how the statistical model generates those observations at each parameter value; the posterior distribution describes uncertainty conditional on the observations and that model. A physical parameter can have one fixed value while its value remains uncertain to us. Assigning it a probability distribution represents that uncertainty under the chosen model, rather than requiring the physical parameter to change randomly between measurements.

For example, suppose ten Bernoulli trials are independent conditional on an unknown heads probability $\theta$, with a beta distribution prior $\mathrm{Beta}(2,2)$. Eight heads and two tails give the posterior distribution $\mathrm{Beta}(10,4)$ by Beta-binomial conjugacy. Its posterior mean is $10/14=5/7\approx 0.714286$, between the prior mean $1/2$ and observed binomial proportion $8/10$. The Bayesian posterior below computes the whole update, and the credible interval below summarizes its remaining uncertainty. The conjugate update is explained in MIT 18.05, Reading 15: beta and normal conjugate priors.

The useful feature is that one posterior distribution supports several questions: a posterior mean estimates a value, a credible interval reports a range, and a posterior predictive distribution predicts a future observation while averaging over parameter uncertainty. With conditionally independent new data, today's posterior distribution can become tomorrow's prior distribution. These conclusions remain conditional on the statistical model and prior distribution; a precise update does not establish that either assumption fits the experiment.

A historical starting point is Thomas Bayes's essay in the 1763 volume of the Royal Society's Philosophical Transactions, communicated after his death by Richard Price. Its opening problem asks for uncertainty about an unknown event probability given observed successes and failures: the same direction of reasoning illustrated by the coin example. See Bayes and Price, An Essay towards solving a Problem in the Doctrine of Chances, opening letter and problem.

For a finite measurable partition $(B_j)$ with positive probabilities, and an event $A$ of positive probability,Its beauty is that a model predicting observations from a hidden state can also describe uncertainty about the state after an observation. Each numerator is a joint probability; the denominator, supplied by the law of total probability, makes the answers sum to one. This is a direct consequence of conditional probability, rather than an extra axiom.

Select the coin from law of total probability once, then toss that same coin repeatedly, with conditionally independent tosses given its identity. Let $B$ denote the $3/4$-head coin, initially of prior probability $1/5$. After one head $H$, its posterior probability isThe likelihood of a head is larger under $B$, but $B$ remains less likely than the other coin because its prior probability was smaller. A large likelihood for an observation does not itself imply a large posterior probability for the hypothesis.

After an ordered sequence containing $h$ heads and $t$ tails, its likelihood ratio between the two coins is $(3/2{)}^h(1/2{)}^t$. Bayes' theorem therefore gives posterior oddsEach head multiplies the odds by $3/2$ and each tail by $1/2$. Three heads give posterior probability $27/59$, still below one half; four give $81/145$, above one half. The finite-model update and the distinction between prior probability, likelihood and posterior probability are set out in MIT 18.05, Reading 11: Bayesian Updating with Discrete Priors.

The hidden selection also creates dependence. Each toss has marginal head probability $11/20$, but two heads have joint probability $(4/5)(1/2{)}^2+(1/5)(3/4{)}^2=5/16$, greater than $(11/20{)}^2=121/400$. Their indicator covariance is $1/100$. Seeing a head makes the head-favoring coin more likely, so it also raises the chance of a future head. This is covariance induced by a shared latent variable, despite conditional independence of the tosses given the coin.

Historically, Thomas Bayes's essay on an unknown success probability appeared in 1763 after his death, communicated by Richard Price. Its opening problem asks how observed successes and failures constrain that unknown probability; Price's introductory letter explains the connection to reasoning from experience. That is a more substantial inference problem than simply rearranging today's elementary identity. ^[ref]

Give Alice the first qubit $A$ and Bob the second $B$. Compare the Bell state $|{\Phi }^{+}⟩$ with a preparation that tosses a fair classical bit and gives both parties its value. Their density operators, written in the computational basis ordered $00,01,10,11$, areThey have identical diagonal quantum state populations. The off-diagonal entries of the Bell state retain quantum coherence between $00$ and $11$; the classical mixture has none in this basis.

If both parties perform a Pauli measurement of $Z$, the Born rule gives $P(00)=P(11)=1/2$ and $P(01)=P(10)=0$ for both preparations. Perfect agreement here therefore cannot establish entanglement: an ordinary shared random bit already explains it.

Now both measure $X$, whose normalized eigenstates are $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. Changing basis givesThe Bell state again has matching outcomes, each with probability $1/2$. For either product state $|00⟩$ or $|11⟩$, each of the four $X$ outcome pairs has probability $1/4$, so their classical mixture does too. Equivalently, the Pauli correlator $⟨X\otimes X⟩$ is one for the Bell state and zero for this mixture, while $⟨Z\otimes Z⟩$ is one for both.

Taking a partial trace gives ${\rho }_A={\rho }_B=I_2/2$ for both preparations. Thus every measurement confined to Alice's system has the same statistics in the two cases, and so does every measurement confined to Bob's. They reveal the difference by comparing records through classical communication. Conditional on Alice's result, Bob can predict a matching $X$ result for the Bell state; without that result, his outcomes remain evenly split. An uncommunicated local quantum channel, including a measurement with its outcome discarded, leaves his reduced density matrix unchanged. This is quantum no-signalling, even when conditional states change. ^[ref]

The Von Neumann entropy makes the joint-versus-local distinction numerical. With base-two logarithms, the Bell state has eigenvalues $1,0,0,0$, giving $S(AB)=0$, whereas the mixture has $1/2,1/2,0,0$, giving $S(AB)=1$. Both have $S(A)=S(B)=1$. Their quantum mutual information $I(A:B)=S(A)+S(B)-S(AB)$ is consequently two and one bits, respectively. The value two is not two readable classical bits in either local qubit: the matching $Z$ records have ordinary mutual information one bit in both cases. Random records can come from a globally pure state. ^[ref]

These two matching-axis measurements distinguish the specified preparations; they alone are not a test of a Bell inequality. A local hidden-variable theory could assign shared random answers separately to these two axes. Bell theorem concerns incompatible correlations across a suitable larger collection of measurement settings. Bell's 1964 paper established the obstruction for local hidden-variable descriptions of the predicted correlations. ^[ref]

A Bell test compares experimental probabilities or expectation values with a Bell inequality. In a two-setting photon polarization test, each station chooses a polarizer orientation and records an outcome. A coincidence circuit identifies candidate pairs using an agreed timing rule. The choice of which trials enter the statistic is part of the experiment: correlations calculated only from detected pairs can differ from correlations of all emitted pairs. The Polarization CHSH count model separates ideal expectation values from this experimental selection problem.

For a positive random variable $X$, let $D\in \{1,\dots ,9\}$ be its first nonzero base-ten digit. The Benford law is the discrete probability distribution with probability mass functionHere $X$ need not be an integer: $0.0042$, $4.2$ and $4200$ all have $D=4$. Multiplication by an integer power of ten changes the decimal point's position and leaves the first nonzero digit unchanged. Zero has no first nonzero digit, so exclude it. For negative inputs one may explicitly use their absolute values, still excluding zero; that convention must be stated rather than applying a logarithm to a negative number.

The Benford law favors small digits: $\mathbb{P}(D=1)={\log }_{10}2\approx 0.301029996$, whereas $\mathbb{P}(D=9)={\log }_{10}(10/9)\approx 0.04575749$. Thus the model assigns about $30.10\%$ to digit one and $4.58\%$ to digit nine. The masses are positive and their normalization is a finite telescoping series:

Why can digit one be so common? In a decimal significand interval $[1,10)$, the digit-one region $[1,2)$ occupies a larger fraction of the logarithmic scale than the digit-nine region $[9,10)$. The precise assumption is that the fractional part of the base-ten logarithm is distributed uniformly on $[0,1)$; Benford law from logarithmic uniformity turns those unequal interval lengths into probabilities. This makes a surprising digit pattern into a simple geometry of multiplicative size. The logarithmic explanation is also presented in Peter Sheridan Dodds, Benford's Law, essential story.

The Benford law is useful as a proposed model for leading digits when the mechanism supports logarithmic uniformity or an appropriate limiting argument, such as Benford frequencies for powers of an integer. Merely spanning many powers of ten does not establish that assumption. A discrepancy from the Benford law can prompt investigation of the model and data collection; by itself it does not establish fraud, and agreement does not establish authenticity.

Simon Newcomb's 1881 paper, Note on the Frequency of Use of the Different Digits in Natural Numbers, appeared in the American Journal of Mathematics 4, pp. 39–40. Its opening connects unequal leading-digit frequencies with the faster wear of early pages in logarithm tables. The publication metadata and original opening excerpt are available in Newcomb (1881), original publication record and ^[ref]. Frank Benford's The Law of Anomalous Numbers, Proceedings of the American Philosophical Society 78 (1938), pp. 551–572, likewise discusses worn logarithm tables on p. 551 and tabulates observations on p. 553. See Benford (1938), transcription of the original paper with page numbers and publication metadata. These observations motivated study of the digit pattern; they do not make the Benford law universal.

For a positive random variable $X$, write its base-ten logarithm as an integer part plus a fractional part:The floor function handles values below one as well: for $X=0.0042$, $K=-3$ and the decimal significand is $10^Z=4.2$. Since $10^Z\in [1,10)$, the first nonzero decimal digit is $D=\lfloor 10^Z\rfloor$. The monotonicity of the logarithm gives the exact event equivalenceIf $Z$ has the uniform distribution on $[0,1)$, the probability of this interval is its length. Consequently $\mathbb{P}(D=d)={\log }_{10}(d+1)-{\log }_{10}d$, proving the Benford law. This argument assumes uniformity of the fractional part of the logarithm, rather than uniformity of $X$ itself. See the logarithmic-coordinate explanation in Peter Sheridan Dodds, Benford's Law, essential story.

An exact constructed example is the log-uniform distribution on $[1,10)$, with probability density function $f_X(x)=1/(x\ln 10)$ there and zero elsewhere. Its total mass is ${\int }_1^{10}dx/(x\ln 10)=1$. The change of variables formula with $x=10^z$ gives $f_Z(z)=f_X(10^z)(\ln 10)10^z=1$ for $0\le z<1$. Equivalently, integrating directly givesFor example, the probability of first digit four is ${\log }_{10}(5/4)\approx 0.096910013$. By contrast, an ordinary uniform distribution for $X$ on $[1,10)$ has probability density function $1/9$ there, so $\mathbb{P}(D=d)={\int }_d^{d+1}dx/9=1/9$ for every digit. This is a concrete counterexample to any claim that the Benford law holds for every positive random variable.

More generally, a log-uniform distribution on $[A,B)$ with $0<A<B$ has uniform $Y={\log }_{10}X$ on $[{\log }_{10}A,{\log }_{10}B)$. If its logarithmic span ${\log }_{10}(B/A)=m$ is a positive integer, reduction to the fractional part wraps an interval of length $m$ around the unit interval exactly $m$ times, making $Z$ uniform even when $A$ is not a power of ten. Arbitrary partial-decade truncations generally do not have this property: for a log-uniform distribution on $[1,2)$, the first digit is always one.

Under the uniform fractional-logarithm assumption, multiplication by any constant $c>0$ replaces $Z$ by $Z^{\prime }=\{Z+{\log }_{10}c\}$. This is a rotation modulo one, which preserves the uniform distribution and therefore preserves the Benford law; see scale invariance of decimal significands. For the constructed $[1,10)$ example, multiplying by $0.001$ gives a log-uniform distribution on $[0.001,0.01)$ with the same digit probabilities. The claim concerns the decimal significand law: there is no proper probability density function proportional to $1/x$ over all of $(0,\infty )$, since its integral diverges. The existing scale-invariant prior describes that unnormalizable measure, while a finite log-uniform distribution is normalizable.

A Bernoulli process is a sequence of independent and identically distributed random variables with a Bernoulli distribution: $\mathbb{P}(B_n=1)=p$ and $\mathbb{P}(B_n=0)=1-p$. It models repeated Bernoulli trials with a fixed success probability and no dependence between trials. The count $K_n={\sum }_{j=1}^n{B}_j$ has a binomial distribution, expected value $np$, and variance $np(1-p)$; the first success time has a geometric distribution on $\{1,2,\dots \}$. The observations $B_n$ and the accumulated count $K_n$ are different stochastic processes: the observations are independent, while consecutive counts share all earlier observations. ^[ref]

Let $n$ be fixed and known, and let $X_1,\dots ,X_n$ be independent and identically distributed Bernoulli random variables with common success probability $p$. Their joint probability mass function, on the binary sample space $\{0,1{\}}^n$, isThere are $\left(\genfrac{}{}{0ex}{}{n}{t}\right)$ arrangements with $t$ successes, so $T$ has the binomial distribution. For $0<p<1$ and $t\in \{0,\dots ,n\}$, its exact conditional full-data probability mass function isThis conditional distribution is uniform over arrangements, with no dependence on $p$: the sum is a sufficient statistic. ^[ref]

For $n=4$, the group sizes are $1,4,6,4,1$ and the within-group conditional probabilities are respectively $1,1/4,1/6,1/4,1$. For example, $T=2$ leaves exactly $0011,0101,0110,1001,1010,1100$, each with conditional probability $1/6$. At $p=1/4$, each such sequence has unconditional probability $9/256$ and the group has probability $54/256$. At $p=1/2$, these values are $1/16$ and $6/16$; at $p=3/4$ they return to $9/256$ and $54/256$. The conditional weights stay $1/6$ in all three cases.

At the boundary $p=0$, only $T=0$ has positive probability; at $p=1$, only $T=n$ does. Dividing by $P_p(T=t)$ on the other groups would divide by zero. Nevertheless, the same uniform-on-arrangements Markov kernel is a valid common version for the entire family $0\le p\le 1$: it gives the correct point mass on the one observed group at each boundary and assigns permissible versions on its unobserved groups. Define the joint probability mass function there by the original product of the one-trial Bernoulli probability mass functions, or use the limiting convention $0^0=1$ in the compact formula.

The sample mean $T/n$ is an unbiased estimator of $p$ and retains the sufficient statistic because $n$ is known. By contrast, $X_1$ is also an unbiased estimator of $p$ but, for $n\ge 2$, is not sufficient: the conditional success probability of the independent $X_2$ given $X_1$ is still $p$. Thus unbiasedness does not imply sufficiency.

Sufficiency also does not imply completeness of a statistic. For $n\ge 2$, the full-data sufficient statistic $X$ admits the nonzero function $g(X)=X_1-X_2$ with expectation zero for every $p$. Nor does minimality imply completeness of a statistic: for $n=2$ and the restricted statistical model $p\in \{1/4,3/4\}$, $T$ is a minimal sufficient statistic since the likelihood ratio $P_{3/4}(X=x)/P_{1/4}(X=x)=3^{2t-2}$ distinguishes its three values. Yet $g(0)=3,g(1)=-5,g(2)=3$ has expectation $(27-30+3)/16=0$ at both parameter values and is never zero. Changing the parameter family can change completeness of a statistic even while preserving sufficiency.

A Bernoulli string is a finite sequence of independent binary digits, each taking value one with a specified probability. Unless stated otherwise, all positions use the same probability.

Take an IID random variable source of bits with $P(X_j=1)=1/4$. Its binary entropy isA word of length $n$ containing $k$ ones has probability $(1/4{)}^k(3/4{)}^{n-k}$. Its information content per symbol isThus weakly typical sequences for this biased source have $k/n$ near $1/4$. With tolerance $\epsilon =0.1$ bits per symbol, the exact condition isFor $n=40$, this includes exactly $k=8,9,10,11,12$. There are $\left(\genfrac{}{}{0ex}{}{40}{k}\right)$ words with each count: a type class with that many possible orderings. Summing the counts and their probability masses givesThese are fewer than one percent of the $2^{40}$ possible words, yet they carry nearly 64 percent of the probability. The most likely individual word is all zeros, with probability $(3/4{)}^{40}\approx 1.006\times 10^{-5}$; it is outside this typical set. No contradiction is involved: the type classes near ten ones each contain many more individual words.

For a function $f:[0,1]\to \mathbb{R}$ and an integer $n\ge 1$, its Bernstein polynomial isThe Bernstein basis supplies nonnegative weights whose sum is one. Equivalently, the weights are the probabilities of a binomial distribution, and the result is an expected value. At a given $x$, this averages the sampled values $f(k/n)$ with weights concentrated near $k/n=x$.^[ref]

The endpoints are exact: $B_{n}f(0)=f(0)$ and $B_{n}f(1)=f(1)$. Interior sampled points need not be interpolated. For example, binomial distribution moments give $B_n(x)=x$ and $B_n(x^2)=x^2+x(1-x)/n$. Thus every affine function is reproduced exactly, whereas $x^2$ is generally changed. Bernstein polynomial degree preservation is a statement about degree of a polynomial, not equality of the entire polynomial.

For a continuous $f$, the uniform continuity of $f$ and the variance $\mathrm{Var}(K/n)=x(1-x)/n\le 1/(4n)$ explain uniform convergence. Split the expected value of the error into $|K/n-x|<\delta$, where uniform continuity makes the change small, and its complement, whose probability is at most $1/(4n{\delta }^2)$ by Chebyshev's inequality. The bounded function $f$ bounds the remaining error. This proves convergence without assuming differentiability.

For $f\in C([0,1{]}^d)$, the tensor-product Bernstein polynomial is the expected value of $f(X_1/n,\dots ,X_d/n)$ for independent random variables $X_j\sim \mathrm{Bin}(n,x_j)$. The same uniform continuity and coordinatewise variance estimates give uniform convergence for each fixed finite $d$.

A Berry phase is the geometric phase acquired when a nondegenerate energy eigenstate follows a closed cycle of parameters of a Hamiltonian operator in the ideal limit of the adiabatic theorem. The energy eigenstate must stay separated from other energy eigenstates by a nonzero spectral gap, and the parameters must change sufficiently slowly. Returning the Hamiltonian operator to its starting value returns the state to the same ray, but its quantum phase can remember the journey. The total quantum phase contains both the dynamical phase and the Berry phase.

The beauty of the Berry phase is that, within the adiabatic theorem's limit, it depends on the oriented path of the parameters rather than how long the journey takes. For a spin one-half qubit, the Berry phase of a Bloch-sphere latitude turns that memory into half a solid angle. Berry-phase interference with a reference path converts this geometry into a detection probability.

Michael Berry's 1984 paper formulated the geometric phase for cycles governed by the adiabatic theorem and derived the solid angle result for spin. Its bibliographic details are Proceedings of the Royal Society of London A, volume 392, pages 45–57, DOI 10.1098/rspa.1984.0023. ^[ref]; ^[ref]. The wider concept of geometric phase does not require every cycle to obey the adiabatic theorem: Aharonov and Anandan formulated a geometric phase for general cyclic quantum evolution in 1987. ^[ref].

An isolated qubit that acquires only a global phase has unchanged measurement probabilities. To reveal the Berry phase, use two coherent paths of a Mach-Zehnder interferometer, represented by a path qubit. In one arm the internal spin one-half qubit follows the loop; in the reference arm it stays in its starting energy eigenstate. Assume equal arm amplitudes, identical final internal states, cancelled relative dynamical phases, and calibrated fixed apparatus phases. In the model above, hold the reference Hamiltonian operator at its initial value for the same duration. Both arms then have energy eigenvalue $-\Delta /2$ and acquire the same dynamical phase $\Delta T/(2\hslash )$, so their relative dynamical phase vanishes in the ideal limit of the adiabatic theorem.

The joint quantum state before recombination then factors asThe Berry phase is now a relative quantum phase between the two paths. The calibrated plus output projects onto $|+⟩=(|\mathrm{ref}⟩+|\mathrm{loop}⟩)/\sqrt{2}$. The Born rule givesFor the Berry phase of a Bloch-sphere latitude, the plus-port probabilities at $\theta =0,60^{\circ },90^{\circ },180^{\circ }$ are respectively $1,1/2,0,1$. This ideal Mach-Zehnder interferometer measurement cannot distinguish $\gamma$ from $-\gamma$ or $\gamma +2\pi$ by itself; scanning a known reference phase can resolve the sign. An uncancelled relative dynamical phase would shift the fringe as well.

For real $z$, the probabilities $J_m(z{)}^2$ over integer angular-momentum modes satisfy ${\sum }_m{m}^2{J}_m(z{)}^2=z^2/2$. Differentiate the unit-modulus function $e^{iz\cos \theta }$ and use Parseval to obtain the identity; its zero first moment follows from the symmetry of opposite orders.

The beta distribution is an absolutely continuous probability distribution for a random variable $X$ between zero and one. It can describe uncertainty about a probability, such as an unknown coin's chance of heads. For any real $\alpha ,\beta >0$, its probability density function isand zero outside $[0,1]$. The normalizer is the Euler beta function, related to the gamma function by the beta-gamma identity:Thus the probability density function integrates to one. ^[ref].

The endpoint behavior explains why the restrictions are exactly $\alpha ,\beta >0$: near zero the probability density function behaves like a constant times $x^{\alpha -1}$, and near one like a constant times $(1-x{)}^{\beta -1}$. These improper integrals converge for positive parameters, even when the probability density function diverges at zero for $\alpha <1$ or at one for $\beta <1$. There is no probability mass at either endpoint: $\mathbb{P}(X=0)=\mathbb{P}(X=1)=0$. A probability density function's height is not a probability; an interval's probability is the area under it.

Some useful examples of the beta distribution are:

The arcsine distribution curve is plotted only for $0.001\le x\le 0.999$; its finite plotted ends do not represent finite endpoint values of the probability density function. Using the inverse sine, the shaded central interval has exact probability $\mathbb{P}(0.1<X<0.9)=1-4\arcsin (\sqrt{0.1})/\pi \approx 0.5903$, calculated analytically rather than by treating the cropped curve as normalized.

To separate location from spread, write $\kappa =\alpha +\beta$ and $m=\alpha /\kappa$. The mean and variance of the beta distribution areThe mean specifies the balance between the parameters; their sum $\kappa$ controls concentration about that mean. Multiplying both parameters by the same positive number preserves the mean and changes the variance. Larger $\kappa$ puts more area near $m$ and makes the probability density function taller there; the total area stays one. Small $\kappa$ allows much greater spread, sometimes favoring the endpoints rather than the mean itself.

For example, keeping mean $m=0.3$ and taking $\kappa =5,20,80$ gives the beta distributions $\mathrm{Beta}(1.5,3.5)$, $\mathrm{Beta}(6,14)$, and $\mathrm{Beta}(24,56)$. Their variances are $0.035$, $0.01$, and $0.21/81\approx 0.00259259$, respectively. Parameters need not be integers.

The formulas follow from two short moment calculations. Multiplying the probability density function by $x$ or $x^2$ increases the first parameter of the Euler beta function:Apply the Gamma function recurrence to the beta-gamma identity, then subtract $(\mathbb{E}[X]{)}^2$ from $\mathbb{E}[X^2]$ to obtain the variance.

The beta distribution is especially convenient as a prior distribution for the unknown success probability $P$ of Bernoulli trials. The Beta-binomial conjugacy says that $h$ successes and $t$ failures update a prior distribution $\mathrm{Beta}(\alpha ,\beta )$ to a posterior distribution $\mathrm{Beta}(\alpha +h,\beta +t)$. For example, suppose a coin has prior distribution $\mathrm{Beta}(2,2)$ and we observe eight heads and two tails. Its posterior distribution is then $\mathrm{Beta}(10,4)$, with posterior mean $5/7$. ^[ref].

For independent geometric random variables with success probability $p$, observing $d$ terminal events and $F$ preceding failed Bernoulli trials gives likelihood function proportional to $p^d(1-p{)}^F$. Multiplication by a beta distribution prior $\mathrm{Beta}(a,b)$ gives the posterior distribution $\mathrm{Beta}(a+d,b+F)$. With complete waiting times $x_1,\dots ,x_n$, take $d=n$ and $F={\sum }_i(x_i-1)$.

A waiting time that is right-censored after $c$ failed trials contributes $(1-p{)}^c$: increase $F$ by $c$ without increasing $d$. This requires the censoring mechanism to be ignorable, as with an externally fixed observation horizon. Counting a censored waiting time as an observed event changes the likelihood function and biases the update.

For probability distributions $p,q$ on the same finite alphabet, the Bhattacharyya coefficient measures their square-root overlap:The vectors $(\sqrt{p_i}{)}_i$ and $(\sqrt{q_i}{)}_i$ have unit Euclidean norm, so the Cauchy-Schwarz inequality gives $0\le B\le 1$. Equality $B=1$ holds exactly when the distributions agree; $B=0$ holds exactly when they assign positive probability to disjoint outcomes. For $p=(0.9,0.1)$ and $q=(0.6,0.4)$, $B=\sqrt{0.54}+0.2\approx 0.934847$. This overlap is not the probability that two independent draws agree, which here is $0.58$.

The fidelity of commuting mixed states reduces to this coefficient. More generally, the measurement characterization of quantum fidelity minimizes it over outcome distributions of a common measurement in quantum mechanics, connecting classical overlap to the unsquared quantum fidelity.

A biased random walk on the integer line takes unit steps right and left with unequal probabilities $p$ and $1-p$, using independent and identically distributed random variables as increments. For $S_0=0$, the expected value is $n(2p-1)$: positive for $p>1/2$ and negative for $p<1/2$. Changing $p$ to $1-p$ reverses the preferred direction. The exact endpoint probability mass function is given under simple random walk on the integer line.

For example, with $p=0.65$ each step has expected value $0.30$ and variance $0.91$. After $80$ steps,The displacement of the expected value from zero is about $24/8.532\approx 2.813$ standard deviations, yet individual sample paths still move left as well as right, as the figure under simple random walk illustrates. Bias lets this elementary stochastic process model an average directional tendency while retaining fluctuations. Boundary and return questions have their own treatments under gambler's ruin and the specialized children below.